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Home , Ideal gas, Real gas

Topic 4-

I. Behavior of Gases

(Read p. 66-69)

S.Panzarella I. Kinetic Molecular Theory – (under ideal circumstances) - explains how an “ideal” acts Let’s watch The Kinetic Molecular Theory: Properties of Gases (6:09)…http://education- portal.com/academy/lesson/the-kinetic-molecular-theory-properties-of-gases.html#lesson

A. Particles in an … • have no . • have elastic . • are in constant, random, straight-line motion. • don’t attract or repel each other. • have an avg. KE directly related to (KE = ½ MV2) S.Panzarella B. Gas behavior is most ideal… C. Real Gases Particles….

 at low  Real gases deviate from

 at high an ideal gas at low temperatures and high  in nonpolar pressures. atoms/molecules of low • Have their own volume molecular mass ie. H2 (due to increased ) and He • Attract each other (due to increased temps  think about the conditions for a beach vacation

S.Panzarella Homework

Read page 66, 68-69 pg 4-5 of guide, #’s 1-11

S.Panzarella Guide pg 4-5 #1-11

1. C 2. C 3. D 4. C 5. B 6. C 7. B 8. Real gases have a small, but significant volume or they have weak IMF’s 9. A 10. A 11. A

S.Panzarella CHEM DO review – You try 1. Which gas will most closely 5. Under the same conditions of resemble an ideal gas at STP? temperature and pressure, which of the 1) SO2 2) NH3 3) Cl2 4) H2 following gases would behave most like an ideal gas? 2. At STP, which gas would most likely (1) He(g) (2) NH3(g) behave as an ideal gas? (1) H2 (2) CO2 (3) Cl2 (4) SO2 (3) Cl2(g) (4) CO2(g)

3. Which gas has properties that are most similar to those of an ideal gas? 6. Which gas has properties that are most (1) O2 (2) H2 (3) NH3 (4) HCl similar to those of an ideal gas?

4. Under which conditions does a real (1) N2 (2) O2 (3) He (4) Xe gas behave most like an ideal gas? 7. One reason that a deviates (1) at high temperatures and low from an ideal gas is that the molecules of pressures the real gas have (2) at high temperatures and high pressures (1) a straight-line motion (3) at low temperatures and low (2) no net loss of on pressures (4) at low temperatures and high (3) a negligible volume pressures (4) forces of attraction for each other

Chem Review Do Answers

1. 4 2. 1 3. 2 4. 1 5. 1 6. 3 7. 4

S.Panzarella III. The

 Shows the relationships between TEMPERATURE, PRESSURE, VOLUME and number of moles of a gas

 Used to determine what effect changing one of those variables will have on any of the others.

S.Panzarella Bill Nye Video

S.Panzarella Normal conditions……

• Standard Temperature & *Found in Pressure - *STP Reference Table A • 273 K (or 0◦C) and 101.3 kPa or 1 ATM

K = ºC + 273 • Temperature MUST be (KELVIN) when working with gases.

1 Atm = 760 torr =760 mmHg = 101.3 kPa = 14.7 lb/in2

S.Panzarella

1. Combined Gas law Combines 3 gas laws # moles are held constant P, T and V change  Let’s watch (1:05): http://www.youtube.com/watch?v=t-Iz414g ro&feature=player_embedded Formula: found on Table T

P1V1 P2V2

T = T 1 2 S.Panzarella

Combined Gas Law – Table T

P1V1 P2V2 = T1 T2

P1V1T2 = P2V2T1

S.Panzarella COMBINED Gas Law Problem #1

 2.00 L sample of gas at 300. K and a pressure of 80.0 kPa is placed into a 1.00 L container at a pressure of 240. kPa. What is the new temperature of the gas? GIVEN: :

V1 = 2.00 L P1V1T2 = P2V2T1 P1 = 80.0 kPa (80.0 kPa)(2.00 L)(T2) T1 = 300 K V = 1.00 L 2 = (240. kPa) (1.00 L) (300 K) P2 = 240. kPa T2 = 450 K T2 = ?

S.Panzarella COMBINED Gas Law Problem #2

A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.

GIVEN: WORK: 3 V1 = 7.84 cm P1V1T2 = P2V2T1 P1 = 71.8 kPa (71.8 kPa)(7.84 cm3)(273 K) T1 = 25°C = 298 K V = ? 2 = (101.3 kPa) V2 (298 K) P2 = 101.3 kPa 3 V2 = 5.09 cm T2 = 273 K

S.Panzarella 2. Boyle’s Law

 Video: Let’s watch (1:01): http://www.youtube.com/watch?feature=player _embedded&v=DcnuQoEy6wA

 Pressure vs. Volume (Constant Temperature): (think: squeezing a balloon)

 As pressure is INCREASED, volume is DECREASED

S.Panzarella Boyle’s Law con’t.

 INVERSE relationship:

P  PV = K

 formula V

P1V1= P2V2

S.Panzarella BOYLE’S Law Problems

A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.

GIVEN: P V WORK:

V1 = 100. mL P1V1T2 = P2V2T1 P = 150. kPa 1 (150.kPa)(100.mL)=(200.kPa)V2 V2 = ? V2 = 75.0 mL P2 = 200. kPa S.Panzarella 3. Charles’ Law

Let’s Watch Video (33 sec) : http://www.youtube.com/watch?v=XHiYKfAmTMc&feature=player_embedded  Volume vs. Temperature (Constant Pressure) (think: hot air balloon)  As temperature is INCREASED, volume is INCREASED

• As the temperature of the water increases, the volume of the balloon increases.

S.Panzarella Charles’ Law con’t.

 DIRECT relationship: V  V/T = K

T  Formula

V1 T2 = V2 T1

S.Panzarella CHARLES’ Law Problems

A gas occupies 5.00L at 36°C. Find its volume at 94°C.

GIVEN: T V WORK:

V1 = 5.00 L P1V1T2 = P2V2T1

T1 = 36°C = 309K (5.00 L)(367 K)=V2(309 K) V2 = ? V2 = 5.94 L T2 = 94°C = 367K S.Panzarella 4. Gay-Lussac’s Law

 Temperature vs. Pressure (Constant Volume)- (think: car tires or pressure cooker)  As temperature is INCREASED, pressure is INCREASED

S.Panzarella Gay-Lussac’s Law con’t

 DIRECT relationship:  P/T = K P  Formula:

T

P1T2 = P2T1

S.Panzarella GAY-LUSSAC’S Law Problem

 A 10.0 L sample of gas in a rigid container at 1.00 atm and 200. K is heated to 800. K. Assuming that the volume remains constant, what is the new pressure of the gas? GIVEN: P T WORK:

P1 = 1.00 atm P1V1T2 = P2V2T1

T1 = 200 K (1.00 atm) (800 K) = P2(200 K)

P2 = ? P2 = 4.00 atm T2 = 800 K S.Panzarella Homework

See guide pg 5-6 • Part A (use your RB) • Part B #12-21 • Quiz on Thursday

S.Panzarella Topic 4 RB ANSWERS

47) 1 55) 2 48) 3 56) 2 49) 1 57) 3 50) 4 58) 3 51) 1 59) 4 52) 3 60) 4 53) 3 61) 3 54) 1 62) 2 63) 3

S.Panzarella 2 more Gas laws

(TEXT BOOK p. 350-353) Read these pages first!

S.Panzarella 5. Graham’s Law

 States that the rate of effusion (diffusion) of a gas is inversely proportional to the square root of the gas’s .

effuses (and diffuses) nearly three times faster than at the same temperature.

S.PanzarellaS.Barry

Graham’s Law cont.

Gases of SMALL MASS (or ) diffuse faster than gases of higher molar mass.

ex. At STP, which gas will diffuse more rapidly? Use Table S a) He b) Ar c) Kr d) Xe

S.PanzarellaS.Barry 6. Dalton’s Law

 At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases.  Formula Ptotal = P1 + P2 + P3 ...

 Three gases are combined in container T

S.Panzarella Dalton’s Law example #1

gas is collected over water at 22.5°C and 2.72 kPa. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.

GIVEN: WORK:

PH2 = ? Ptotal = PH2 + PH2O P = 94.4 kPa total 94.4 kPa = PH2 + 2.72 kPa PH2O = 2.72 kPa PH2 = 91.7 kPa

S.Panzarella Dalton’s Law example #2  Ex. Air contains , nitrogen, and

and other gases. What is the of O2 at 101.3 kPa of pressure if the PN2 = 79 kPa , the PCO2 = 0.034 kPa and Pothers = 0.95 kPa?

GIVEN: WORK: PO2 = ? Ptotal = PO2 + PN2 + PCO2 + Poth PN2 = 79 kPa 101.3 kPa = PO2 + 79 kPa + 0.034 kPa + 0.95 kPa PCO2 = 0.034 kPa 101.3 kPa = PO2 + 79.984 kPa Pothers = 0.95 kPa 21.316 kPa = PO2 Ptotal = 101.3 kPa S.Panzarella III. Avogadro’s Hypothesis

 Let’s watch http://education-portal.com/academy/lesson/molar-volume-using- avogadros-law-to-calculate-the-quantity-or-volume-of-a-gas.html#lesson

 Definition: Equal volume of gases at the same temperature and pressure contain equal numbers of particles (molecules) regardless of their mass

Let’s watch (2:05) http://www.youtube.com/watch?v=fexEvn0ZOpo&feature=player_embedded S.Panzarella 1) Which rigid cylinder contains the same number of gas molecules at

STP as a 2.0-liter rigid cylinder containing H2(g) at STP?

(a) 1.0-L cylinder of O2(g) (b) 2.0-L cylinder of CH4(g) (c) 1.5-L cylinder of NH3(g) (d) 4.0-L cylinder of He(g)

2) Which two samples of gas at STP contain the same total number of molecules?

(a) 1 L of CO(g) and 0.5 L of N2(g) (b) 2 L of CO(g) and 0.5 L of NH3(g) (c) 1 L of H2(g) and 2 L of Cl2(g) (d) 2 L of H2(g) and 2 L of Cl2(g)

3) At the same temperature and pressure, which sample contains the

same number of moles of particles as 1 liter of O2(g)?

(a) 1 L Ne(g) (c) 0.5 L SO2(g) (b) (b) 2 L N2(g) (d) 1 L H2O(ℓ)

B. :

 The number of molecules in 22.4 L of any gas at STP has been chosen as a standard unit called 1

 1 mole = 22.4 L of any gas at STP contains 6.02x1023 particles

 22.4 L of any gas at STP is = to its molecular mass

S.Panzarella

Molar volume cont.

 Density = molecular mass of gas volume (22.4 L)

 ex. What is the density of 1 mole of oxygen gas?

 ex. Which gas has a density of 1.70 g/L at STP? a) F2 b) He c) N2 d) SO2

S.Panzarella IV.

Recall - Avogadro’s law, which is derived from this basic idea, says that the volume of a gas maintained at constant temperature and pressure is directly proportional to the number of moles of the gas PV=nRT

Video: http://education-portal.com/academy/lesson/the-ideal-gas-law-and-the- gas-constant.html#lesson S.Panzarella

Ideal Gas Law video http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of- changes-to-a-gas.html#lesson • Merge the Combined Gas Law with Avogadro’s Principle: PVV = Rk nTnT UNIVERSAL GAS n is the CONSTANT number of R=0.0821 Latm/molK moles of the R=8.315 dm3kPa/molK gas You don’t need to memorize these values!

S.Panzarella Ideal Gas Law Problem #1

Calculate the pressure in of 0.412 mol of He at 16°C & occupying 3.25 L. GIVEN: WORK: P = ? atm PV = nRT n = 0.412 mol P(3.25)=(0.412)(0.0821)(289) T = 16°C = 289 K L mol Latm/molK K V = 3.25 L P = 3.01 atm Latm/molK R = 0.0821 S.Panzarella Ideal Gas Law Problem #2

Find the volume of 85 g of O2 at 25°C and 104.5 kPa.

GIVEN: WORK:

V = ? 85 g 1 mol = 2.7 mol 32.00 g n = 85 g = 2.7 mol PV = nRT T = 25°C = 298 K (104.5)V=(2.7) (8.315) (298) P = 104.5 kPa kPa mol dm3kPa/molK K R = 8.315 dm3kPa/molK V = 64 dm3 S.Panzarella